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3 years ago

Two number cubes are rolled. What is the probability that the sum of the numbers rolled is either 4 or 8?

Mathematics

1 answer:

sveticcg [70]3 years ago

8 0

There are 6 numbers on each die.

The total number of out comes is 6 x 6 = 36.

To get a sum of 4, the combinations are: 1 & 3, 2 & 2 or 3 & 1.

To get a sum of 8, the combinations are: 1 & 7, 2 & 6, 3 & 5, 4 & 4, 7 & 1, 6 & 2, 5 & 3

The total of those combinations are: 3 + 7 = 10 total combinations of a sum of either 4 or 8.

The probability of either is 10/36, which can be reduced to 5/18.

You might be interested in

Factorize x squared - 1x +10 =0

Natalka [10]

Answer:

no solution exists:

Step-by-step explanation:

x²−1x+10=0

Step 1: Simplify both sides of the equation.

x²−x+10=0

Step 2: Subtract 10 from both sides.

x²−x+10−10=0−10

x²−x=−10

Step 3: The coefficient of -x is -1. Let b=-1.

Then we need to add (b/2)^2=1/4 to both sides to complete the square.

Add 1/4 to both sides.

x²−x+ 1/4=−10+ 1/4

x²−x+ 1/4=−39 /4

Step 4: Factor left side.

(x -1/2)² = −39 /4

Step 5: Take square root.

x −1 /2 =±√ −39 /4

Step 6: Add 1/2 to both sides.

x =−1/2 + 1/2 = 1/2 ±√ -39/4

x = 1/2 ±√ -39/4

No real solutions.

6 0

3 years ago

Find the minimum and maximum of f(x,y,z)=x^2+y^2+z^2 subject to two constraints, x+2y+z=4 and x-y=8.

Alika [10]

The Lagrangian for this function and the given constraints is

$L(x,y,z,\lambda_1,\lambda_2)=x^2+y^2+z^2+\lambda_1(x+2y+z-4)+\lambda_2(x-y-8)$

which has partial derivatives (set equal to 0) satisfying

$\begin{cases}L_x=2x+\lambda_1+\lambda_2=0\\L_y=2y+2\lambda_1-\lambda_2=0\\L_z=2z+\lambda_1=0\\L_{\lambda_1}=x+2y+z-4=0\\L_{\lambda_2}=x-y-8=0\end{cases}$

This is a fairly standard linear system. Solving yields Lagrange multipliers of $\lambda_1=-\dfrac{32}{11}$ and $\lambda_2=-\dfrac{104}{11}$ , and at the same time we find only one critical point at $(x,y,z)=\left(\dfrac{68}{11},-\dfrac{20}{11},\dfrac{16}{11}\right)$ .

Check the Hessian for $f(x,y,z)$ , given by

$\mathbf H(x,y,z)=\begin{bmatrix}f_{xx}&f_{xy}&f_{xz}\\f_{yx}&f_{yy}&f_{yz}\\f_{zx}&f_{zy}&f_{zz}\end{bmatrix}=\begin{bmatrix}=\begin{bmatrix}2&0&0\\0&2&0\\0&0&2\end{bmatrix}$

$\mathbf H$ is positive definite, since $\mathbf v^\top\mathbf{Hv}>0$ for any vector $\mathbf v=\begin{bmatrix}x&y&z\end{bmatrix}^\top$ , which means $f(x,y,z)=x^2+y^2+z^2$ attains a minimum value of $\dfrac{480}{11}$ at $\left(\dfrac{68}{11},-\dfrac{20}{11},\dfrac{16}{11}\right)$ . There is no maximum over the given constraints.

7 0

4 years ago

The length of a page in a yearbook is 7 inches. The top margin is 1/2 inch, and the bottom margin in 3/4 inch. What is the lengt

Andrei [34K]

Can you be more specific?

7 0

3 years ago

Kindly answer this one. Thank you.

steposvetlana [31]

Answer:

See below.

Step-by-step explanation: